Let us simplify the question by breaking it into variables.

Let the total number of the candidates be n.
The number of members to be elected =n-r
Thus a voter can vote for 1 of them or 2 of them till n-r in 967 ways.
That is nC1+nC2+nC3+....nC(n-r)=967 ways
The above sum should immediately get you to the formula
nC0+nC1+....nCn=2^n
So we have to get above formula into this form.
Add nC0 to both sides.
nC0+ nC1+nC2+nC3+....nC(n-r)=967+nC0=967+1=968....(I)

Now, let us take the second part.
A voter can vote in 55 different ways to elect (r – 1) candidates by voting in the same manner
So nC1+nC2+...nC(r-1)=55
nC0+nC1+nC2+.....nC(r-1)=55+nC0=55+1=56.....(II)
The other thing one should know is that nC0=nCn and nC1=nC(n-1)
Replace these values in (II)
nCn+nC(n-1)+....nC(n-(r-1))=56
Add equation (I) to above equation
\(nC0+nC1+....nC(n-r)+nC(n-(r-1))+.....nC(n-(r-r))=968+56=1024=2^n.......n=10\)

So total candidates are 10.
10C1+10C2+...10C(r-1)=55
10+45+..=55
So, r-1=2.......r=3

Answer : 10 and 3

B